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Babytud
1 year 5 months ago

@sumit has already explained, I am explaining again if someone needs more clarity.

There is a theorem for real numbers, that is,  If you have a rational number  then there are irrational numbers x arbitrarily close to  , and vice versa.

in other words, every rational number can be approximated by irrational number and vice versa.

Now if I define a piecewise function in which two pieces are never equal then that function is discontinuous on every point, See this-

$f(x)= \begin{cases} 0 & \text{x is rational} \\ 1 & \text{x is irrational} \end{cases}$

This is discontinuous on EVERY point, why?

Take any point either rational or irrational $x_o$. Let $x_o$ be rational. Now $f(x_o) = 0$, and because of theorem above $x_o^{+}$ is irrational. that is $f(x_o^+) = 1$ , similar argument follows if  $x_o$  is irrational. This shows that function is discontinuous at every point. (this is discontinuous at every point because 1 is never equal to 0).

Now what if I define two pieces which are sometimes equal to each other ;)

Lets say two pieces are  x2 + 1 and tan(x), and my function is this-

$f(x)= \begin{cases} x^2+1 & \text{x is irrational} \\ tan(x) & \text{x is rational} \end{cases}$

As Sumit already exlain that x2 + 1 and tan(x) are definitely equal to each other on exactly 4 points in interval $[0, 4\pi]$, I am directly using this fact here.

For me it doesn't matter that these points are rational or irrational. Be it anything function is continuous on these points.

say one of the point is  $x_o$ means   $x_o$ is a solution of $x^2+1 = tan(x)$

say   $x_o$ is irrational then  $f(x_0) = x_o^2+1$

and $f(x_o^+)=tan(x_0^+)$ (as   $x_o^+$must be rational)

And these two are already equal, then function is continuous on   $x_o$. Similarly on all other points.

Therefore function is continuous on exactly 4 points on given interval.

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KIRAN KUMAR
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